Homework Help:
Theory of relativity - spacetravel

I failed this question on my exam , would appreshiate someone helping me figuring out a correct answer

"Suppose a person on planet X, 10 lightyears away from Earth, sends a readiomessage travelling with the speed of light. An astronomer leaves planet X in a spaceship, at the exact same time as the message is sent, to travel to Earth.

Suppose a second astronomer, recieving the message on Earth, fell down from his radiotelescope three years before he recieved the message.

From the viewpoint of the traveller from planet X: Did the astronomer break his leg before or after the message was sent from planet X? Reason around the two extremecases/options possible.

Reason around the two extremecases/options possible.
Clue: Has to do with traveller relatives speed."

Okay... do you want me to find out at what speed or just want an extreme example? I will do the latter

assume the traveller travels at very low speed ~ 0m/s... so this is a non-relativistic case.. when will he sees the guy broke his legs on earth?... remember signal needs time to travel between two planet, and how does it compare with the time he send out the message?

OK, the other extreme... assume the traveller travels at the speed of light... I'll leave this case for you

Well, in actuality special relativity forbids "time reversile", in other words the order of events cannot be reversed.

The following deviates a little from this problem you posed, but it should help. Suppose a cause happens at point 1 (x1,t1), and its "effect" occurs at point 2 (x2,t2). It is possible using Lorentz Transformation to show that,
t'2 - t'1 > 0; that is, O' can never see the "effect" coming before the "cause".

The lorentz transformation for time between planet X and the spacecraft astronomer is

[itex]t' = \gamma(t - v x)[/itex]

in natural units where [itex]c = 1[/itex]. The second astronomer falls when [itex]t = 7[/itex] years and [itex]x = 10[/itex] light years. The message was sent from planet X when [itex]t = t' = 0[/itex]. The time of the fall event in the spacecraft frame is thus

[itex]t' = \frac{7 - 10v}{\sqrt{1 - v^2}}[/itex].

As you can see from the equation, [itex]t' > 0[/itex] when [itex]v < 0.7[/itex]. So above 70 % the speed of light, the astronomer breaks his leg before the message was sent. If you are concerned about optical delay due to the finite speed of light, you need to consider the time of reception [itex]t'_\mathrm{r}[/itex] in the spacecraft frame,

[itex]t'_\mathrm{r} = \gamma(t - v x) + x'[/itex]

The [itex]x'[/itex] term is the time taken for the light from the event to reach the spacecraft in light years.

If you plot this equation you will see that [itex]t'_\mathrm{r} > 0 \ \forall v \in (-1,1)[/itex]. Hence the astronomer on the spacecraft will always see the emission of the message event before the astronomer breaks his leg. However, the two events will be seen to become arbitrarily close in time in the limit that [itex]v\rightarrow c[/itex].